A car takes off at a stop sign to reach a speed of 5 m/s in 3 seconds. Seeing something in the road, the car comes to a stop in 2.2 seconds. What is the car's initial acceleration and then its deceleration?
I need to understand the equation or I can't understand how to solve the answer, please?
Physics Acceleration?
2020-08-30 11:03 pm
回答 (3)
2020-08-30 11:17 pm
Consider the journey of acceleration:
Initial velocity, u = 0 m/s
Final velocity, v = 5 m/s
Time taken, t = 3 s
v = u + at
5 = 0 + a(3)
a = 1.7 m/s²
Consider the journey of deceleration:
Initial velocity, u = 5 m/s
Final velocity, v = 0 m/s
Time taken, t = 2.2 s
v = u + at
0 = 5 + a(2.2)
a = -2.3 m/s²
The answers:
The car's initial acceleration = 1.7 m/s²
Then, the car's deceleration = 2.3 m/s²
Initial velocity, u = 0 m/s
Final velocity, v = 5 m/s
Time taken, t = 3 s
v = u + at
5 = 0 + a(3)
a = 1.7 m/s²
Consider the journey of deceleration:
Initial velocity, u = 5 m/s
Final velocity, v = 0 m/s
Time taken, t = 2.2 s
v = u + at
0 = 5 + a(2.2)
a = -2.3 m/s²
The answers:
The car's initial acceleration = 1.7 m/s²
Then, the car's deceleration = 2.3 m/s²
2020-08-31 12:36 am
Using v = u + at we have:
5 = 0 + 3t
so, t = 1.67 ms⁻²
Again, using v = u + at we have:
0 = 5 + 2.2a
so, a = -5/2.2 => -25/11 = -2.27 ms⁻²
:)>
5 = 0 + 3t
so, t = 1.67 ms⁻²
Again, using v = u + at we have:
0 = 5 + 2.2a
so, a = -5/2.2 => -25/11 = -2.27 ms⁻²
:)>
2020-08-30 11:25 pm
average acc = ΔV/Δt = 5/3 m/s² = 1.67 m/s²
de-acc = ΔV/Δt = –5/2.2 = –2.27 m/s²
de-acc = ΔV/Δt = –5/2.2 = –2.27 m/s²
收錄日期: 2021-04-24 08:03:45
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